$h'(x)=6h(x)$, and $h(1)=5$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $h(x)=6e^{5x-1}$ (Choice B) B $h(x)=5e^{6x-6}$ (Choice C) C $h(x)=e^{6x-6}$ (Choice D) D $h(x)=e^{6x}$
The general solution of equations of the form $h'(x)=kh(x)$ is $h(x)=C\cdot e^{kx}$ for some constant $C$. This can be found using separation of variables. In our case, $k=6$, so $h(x)=C\cdot e^{6x}$. Let's use the fact that $h(1)=5$ to find $C$ : $\begin{aligned} h(x)&=C\cdot e^{6x} \\\\ h(1)&=C\cdot e^{6\cdot 1} \gray{\text{Plug }x=1} \\\\ 5&=C\cdot e^{6\cdot 1} \gray{h(1)=5} \\\\ 5e^{-6}&=C \end{aligned}$ In conclusion, $h(x)=5e^{6x-6}$.